Re: Unknown level of hash

Zhuang Li wrote:
> Yes. I think it's both useful and fun. I was thinking something similar
> to
> @hash{@a} = map{1} @a; 
> 
> But getting "$hash->{E1}->{E2}->...->{En} = 1;" instead of "$hash{E1} =
> 1; ... $hash{En} =1;".
> 
> What I'd really like to do is:
> 
> Given @a = ('E1', 'E2', ..., 'En');
> 	@b = ('K1', 'K2', ..., 'Km');
> 	@c = ('V1', 'V2', ..., 'Vm');
> 
> To get the following in one line:
> $hash->{E1}->...->{En}->{K1} = 'V1';
> $hash->{E1}->...->{En}->{K2} = 'V2';
> ....
> $hash->{E1}->...->{En}->{Km} = 'Vm';
> 

I'll attempt a quess based on S03 & S04 & S09.
( http://dev.perl.org/perl6/synopsis/ )

S09 says that  @nums[dims 0..2]  means  @nums[0;1;2]
S09 also says that "Everything we've said for arrays applies to hashes 
as well ..."
S04 tells how to process several lists in parallel in for-loop.
S03 tells about unary * list-flattening op.

So what about:

     for @b ¥ @c -> $b, $c { $hash->{dims (*@a,$b)} = $c }

ps. I'm not 100% sure if I got that (*@a,$b) right. I want to add $b to 
@a and feed it to dims as one list.

-- 
Markus Laire
<Jam. 1:5-6>